The sum 39

The sum of the first six terms of the arithmetic sequence is 72, and the second term is seven times the fifth term. Find the first term and the AP difference.

Correct answer:

a₁ =  27
d =  -6

Step-by-step explanation:

$$\begin{gathered} s_6=72 \\ {a}_2=7\cdot a_5 \\ {s}_6=6\cdot a_1+d\cdot (1+2+3+4+5) \\ {a}_1+d=7\cdot (a_1+4d) \\ 72=6\cdot a_1+d\cdot (1+2+3+4+5) \\ {a}_1+d=7\cdot (a_1+4d) \\ 72=6\cdot a_1+d\cdot (1+2+3+4+5) \\ {a}_1+d=7\cdot (a_1+4\cdot d) \\ 6a_1+15d=72 \\ 6a_1+27d=0 \\ Row2-Row1\to Row2 \\ 6a_1+15d=72 \\ 12d=-72 \\ d=\frac{-72}{12}=-6 \\ {a}_1=\frac{72-15d}{6}=\frac{72-15\cdot (-6)}{6}=27 \\ {a}_1=27 \\ d=-6 \\ \text{ Verifying Solution: } \\ {a}_2=a_1+d=27+(-6)=21 \\ {a}_3=a_2+d=21+(-6)=15 \\ {a}_4=a_3+d=15+(-6)=9 \\ {a}_5=a_4+d=9+(-6)=3 \\ {a}_6=a_5+d=3+(-6)=-3 \\ s=a_1+a_2+a_3+a_4+a_5+a_6=27+21+15+9+3+(-3)=72 \end{gathered}$$

$d==-6$

Try another example